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Current Question (ID: 16688)

Question:

$\text{A battery of emf } E \text{ and internal resistance } r \text{ is connected to a variable resistor } R \text{ as shown below. Which one of the following is true?}$

Options:

1. $\text{The potential difference across the terminals of the battery is maximum when } R = r.$
2. $\text{The power delivered to the resistor is maximum when } R = r.$
3. $\text{The current in the circuit is maximum when } R = r.$
4. $\text{The current in the circuit is maximum when } R \gg r.$

Solution:

$\text{Hint: The maximum power transfer occurs when the load resistance equals the internal resistance } (R = r).$ $\text{Explanation: The power delivered to the resistor } (P) \text{ is given by:}$ $P = I^2 R \text{ or } P = \frac{E^2 R}{(R+r)^2}$ $\text{To find the condition for maximum power, we differentiate } P \text{ with respect to } R \text{ and set it to zero. This shows that power is maximized when } R = r. \text{ This is known as the maximum power transfer theorem.}$ $\text{The potential difference across the terminals is maximum when } R \rightarrow \infty \text{ (open circuit), not when } R = r.$ $\text{The current } (I) \text{ in the circuit is given by } I = \frac{E}{R+r}. \text{ It is maximum when } R \rightarrow 0 \text{ (short circuit), not when } R = r.$ $\text{The current in the circuit is maximum when } R \ll r, \text{ not when } R \gg r.$ $\text{Hence, option (2) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}